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# Bernoulli's equation (part 3)

Video transcript

Where we left off, we had this
canister, because it had a closed top and it had a vacuum
above the fluid. The fluid on top had an area
of A1, and I poked a little hole with a super-small
area A2. I said that the area of
A2 is so small, it's 1/1,000 of area 1. Then we used the continuity
equation. We said the velocity, the rate
at which the surface is moving up here, V1, times area 1, the
whole surface area of the liquid, has to be equal to the
output velocity, which we're trying to figure out as a
function of everything else times this output area. I made a mistake. I don't know if I did this in
the last video, or I did this is in the mistake video. So we know that the initial top
velocity times this top area is equal to the output
velocity times-- instead of writing area 2, we could write
area 1 over 1,000. You can get rid of the area 1 on
both sides, and then you're saying that the velocity up here
is equal to the rate at which the top of the surface
moves down and is equal to 1/1,000 of the velocity of the
liquid spurting out of this little hole. With that, we actually have the
three variables for the left-hand side of Bernoulli's
equation. What are the variables on
the left-hand side? What is the pressure at this
point where we have a hole? This is an important thin. When we talk about Bernoulli's--
let me rewrite Bernoulli's equation. It's P1 plus rho gh1 plus rho V1
squared over 2 is equal to P2 plus rho gh2 plus rho
V2 squared over 2. We figured out all of these
terms. Now let's figure out the things that we have
to input here. What is the pressure
at point two? This is the important thing. You might want to say, and this
was my initial reaction, too, and that why I made a
mistake, is that what 's the pressure at this depth
in the fluid? That's not what Bernoulli's
equation is telling us. Bernoulli's equation is telling
us actually what is the external pressure
at that hole. When we did the derivation, we
were saying how much work-- this was kind of the work
term, although we played around with it a little bit. But if we look at the water
that's spurting out of the hole, it's not doing any work,
because it's not actually exerting force against anything
so it's not actually doing work. When we think about the
pressure, the output pressure, it's not the pressure at that
depth of the fluid. You should think of it as the
external pressure at the hole. In this case, there is no
external pressure at the hole. Let's say that if we closed
the hole, then at that point, sure. The pressure would be the
pressure that's being exerted by the outside of the canister
to contain the water, in which case, we would end up
with no velocity. The water wouldn't
spurt anywhere. But now we're seeing the
external pressure is zero. That's what the hole essentially
creates. We're going to say that P2 is
zero, so this pressure was zero, because we're
in a vacuum. P2 is also zero, so both
of these are zero. Remember, that's the
external pressure. P1 is the external pressure to
the input to the pipe, and you can view this as a pipe. I could redraw it as a pipe that
looks like it has a big hole on the top, and it goes
down to some level to a super-small hole like that. This would be a vacuum, and the
fluid is just going in and it's spurting out of this end. Anyway, the pressure going into
the pipe is zero, and we said since we put a hole, the
pressure coming out of the pipe is zero, so we're
doing no work. What is this term? This was the potential energy
term, and we said that h1 is equal to h. We're saying that this is zero
height, so now this simplifies to rho times gravity times h
plus rho times V1 squared. V1, we said, is equal to this,
so this is rho over 2 times V2 over 1,000 squared. I just substituted V2
over 1,000 for V1. That equals the pressure at
the hole, the external pressure at the hole, which
is zero, plus h2. This is h2 right here, which
we said is zero. We determined that the hole was
poked at height zero, so this is also zero. That equals this kinetic
energy-like term. It's not exactly
kinetic energy. It's rho times V squared
divided by 2. One thing that we can
immediately see is that we have all these rhos on both
sides of the equation, so we can divide both sides by rho and
get rid of all of those. Then we can multiply both sides
of the equation by 2, and we get 2gh plus V2 squared
over-- what's 1,000 squared-- over 1 million. That is equal to V2 squared. We could do the exact thing. We could subtract 1 over 1
million V2 squared from both sides, and we would get 0.999999
V2 squared, but let's just say for the sake of
simplicity, or let's say, if this wasn't 1,000, but 1
million, and that this surface was much bigger, we see that
this term becomes very, very, very small. If that hole is one millionth
of the surface area, then it becomes really insignificant,
so we can ignore this term because it just makes things
complicated, and we're assuming this is a really,
really large number, and that this hole is much smaller than
the surface area of the fluid. This is like poking a
hole in Hoover Dam. Hoover Dam is backing up this
huge lake, and you poke a hole in it, so that hole is going to
be this very small fraction of the surface area
of the fluid. You can only make this
assumption when that output hole is much smaller than
the input hole. With that said, what is
the output velocity? The velocity-- you just take
the square root of both sides-- is the square
root of 2gh. That is the output velocity. What is the amount of liquid
that flows out per second? We figured that out already. It's a column of fluid that
comes out, so per second, the length of the column of fluid
will be the velocity times time, and then the cross-section
of that column is equal to the output area. If I wanted to know the flow
coming out, the flow coming out or the flux coming out would
be equal to the hole's area times the hole's
output velocity. That would equal the area times
the square root of 2gh. We could use that actually solve
problems in the future if we had actual numbers. I only have a minute
and a half left. I'll see you in the
next video.